text:
A review of $\eta$ efficiency factor for propellers
A glider routinely creates 20x lift force than its propulsive force: the vertical component of gravity, which is sine angle of descent × weight.  Is it's "efficiency" is greater than 1?  Is it 20?  Maybe more? It's also moving forward.  Cdi + Cd0.
Is it possible that propellers, being airfoils, like wings, have an efficiency factor greater than 1?  More like 4?
Let's test this with our Cessna 172R with an engine output torque of 350 footpounds spinning a 4 foot prop at 2700rpm. 180 horsepower.

Thrust = Drag

But wait, which drag?  What comes first?  Engine spins prop.  Engine power output in footpounds = Prop drag in foot pounds.  How much is this?

350 footpounds/0.75 × prop radius = 350/1.5 = 233 pounds of drag

The propeller is an airfoil, just like a wing, with a Lift/Drag ratio.  Multiply by 4 and we have 932 pounds of thrust.
Discussion of "Power" as Force × distance/time for propellers
In the Cessna example 350 foot pounds yields 233 pounds of drag over a larger rotational distance than the engine crankshaft radius. (Imagine the force of the pistons!).
But it is the 233 pounds of drag which produces around 932 pounds of (horizontal lift) thrust.

How do airfoils multiply force?

force × distance/time

Simply put, the leading edge of the airfoil where force is applied to move air out of the way is a smaller distance than the chord of the wing.  Lifting force is created over this chord above and below the wing, in the same amount of time.
Discussion of Thrust = $\eta\frac{P}{V}$
For steady state flight:
Thrust = Drag
$\eta$Power = Thrust × Velocity = Drag × Velocity = 1/2$\rho$V$^2$SCd × V
Thrust × Velocity = 1/2$\rho$V$^3$SCd
Thrust × Velocity = $\eta$Power

$\eta$Power/V = Thrust, Velocity factors out !

What is the efficiency factor?

is true efficiency force of pistons vs force of thrust?

Or even potential energy of fuel vs force of thrust per second, per hour etc.

@sophit, the horsepower dissipated by a falling rock is its weight times its vertical speed- see my post below. -NN

Downvoted because I don't think this site should be the place to learn high-school physics, but a place to learn about aviation. You've been asking questions on basic aircraft physics here for four years and you still haven't learnt the difference between force, energy and power, and which concepts you can apply to each of these quantities.

Downvote because there is no question involved, and definition of efficiency is wrong

@RobertDiGiovanni Anyway I've not yet understood what your problem with the propellers actually is

Re opening sentence-- should read "A glider routinely creates 20x lift force than its propulsive force: the component of gravity that acts parallel to the flight path, which is sine angle of descent × weight. "

Efficiency ($η$) Cannot Exceed 1

Law of conservation of energy prohibits that.

A glider routinely creates 20x lift force than its propulsive force: the vertical component of gravity, which is sine angle of descent × weight. Is it's "efficiency" is greater than 1? Is it 20? Maybe more?

That's not its "efficiency" but the ratio of $\frac {Output\ Force}{Input\ Force}$ which is greater than 1. The same result can be obtained through reduction gears. Given enough leverage, you can obtain a million times more force (torque) than what you started with, but you can never obtain any more power than what you started with.

This reduction gear system may produce a million times greater torque, but it will also inevitably have a million times less rotational speed, such that their product (power) remains the same.

Similarly, a wing may produce much greater lift than drag, but the "lift power" of the wing is:

$\small{Lift\ power = Lift \cdot Velocity \cdot cosθ}$

Since θ is $90°$ (lift is perpendicular to aircraft velocity), cosθ and thus lift power is zero. What is the propulsive efficiency of the wing? Zero - by definition.

Efficiency is not $\frac {Force\ Obtained}{Force\ Invested}$
efficiency is $\frac {Power\ Obtained}{Power\ Invested}$

Propulsive Efficiency ($η$)
This is the efficiency of conversion of power invested (shaft power) into thrust power.
$\large{η = \frac {Thrust\ Power}{Shaft\ Power}}$
Propulsive efficiency can never be greater than 1, since that violates the law of conservation of energy. In practice, it will always be less than 1 since there will always be inefficiencies.
$\small{Thrust\ power = Speed \cdot Thrust}$
$\small{Shaft\ power = Rotation\ Speed \cdot Torque}$
Therefore:
$\large{η = \frac {Speed \cdot Thrust}{Rotation\ Speed \cdot Torque}}$
This is the correct expression for propulsive efficiency. Note that speed here is in TAS.

Force and Power
In the question, it seems like it's being assumed that the L/D ratio of the propeller blades is its "efficiency". While it's true that both L/D ratio and $η$ are indicative of efficiency, both are fundamentally different:
L/D is a ratio of forces, η is a ratio of powers.
Let's understand $η$ graphically. For an engine operating at a constant power of 1 watt and an (unrealistic) propeller with $η = 1$, the thrust-speed graph1 looks like this:

This is the thrust boundary that, for the given power, cannot be exceeded (let's call this the "limit curve"). As speed reduces to zero, thrust that a 1 watt engine can produce tends to infinity. The opposite is true as speed tends to c (speed of light). By definition, 1 watt of power at a speed of 1 $m/s$ yields a thrust of 1 Newton.

Now let's compare this limit curve to the thrust-speed curve of a fixed pitch propeller driven by an engine at a constant power.

It can be seen that the limit curve is never trespassed, as prohibited by the law of conservation of energy.
It was earlier stated that $η = \frac{Thrust\ Power}{Shaft\ Power}$
$\large{η = \frac{T V}{P}}$
$\large{T = \frac {η P}{V}}$
If P and V are fixed, then $η$ only depends on T. Therefore, at a given speed:
$\large{η = \frac{T}{T'}}$
Where T' is the limit curve thrust for that speed.
Dividing T by T' is how we got the graph for $η$. The following features can be observed in this graph:

At $V_0$, $η$ is (always) zero. That is because the maximum thrust that can be produced at $V_0$ (thrust at $η=1$) tends to infinity.

$V_1$ is when $η$ is maximum - around 75% in this case. This is the cruise speed for which the propeller was designed - to provide maximum efficiency.

At $V_2$, $η$ is zero again, because the blades have reached their $α_0$ (zero-lift AoA). The maximum speed that can be attained in level flight for the given engine power is always less than this speed2.

The blades stall at $V_3$. (there will also be a very high speed at which the blades will experience a negative-stall.)

1 This graph is not directly applicable to engines that carry their own propellant (rocket engines).
2 The exact maximum speed can be calculated using math from this answer.

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@AdityaSharma your follow up is outstanding.  I am only starting, so please bear with me.  Excellent blade element theory reference to read.  Here is my observation:  are we really talking "constant power"?  Do we not have to add more fuel to go faster?  For constant RPM, as we increase pitch, prop drag increases, we must increase throttle!  So now efficiency is in its pure form: miles per gallon.  Work continues.

@RobertDiGiovanni Thanks! will surely read that reference. (Also, let's continue this discussion in chat).

This question, which is remarkable, appears to largely relate to a misunderstanding of general theory.  Consequently, we are caught on the horns of a dilemma: To what extent are we, as responsible persons, required to correct the misunderstandings formed of obvious misimpressions regarding the subject at hand?  Understandably, the misinformed person will respond as tho being judged.  Nevertheless, such is not our intent.  We cannot possibly correct those misunderstandings unless the person holding them is willing to set them aside and objectively consider theory from a correct point of view.
The content of this answer is somewhat philosophical - not really engineering theory, not really science, perhaps unnecessary.  But consider the following -

The supreme misfortune is when theory outstrips performance.
$   $ da Vinci

and

Nothing has a performance ratio of 1.

Both of these statements come from a) an observation of human nature and b) an observation of demonstrated fact.  We cannot have a perpetual motion machine.  A glider with a performance ratio of 1 would stay aloft indefinitely.
The performance ratio of a glider is given by its lift to drag ratio.  Consequently a glider with a lift to drag ratio of 50 would have an efficiency of 98 percent. The impossibility is obvious if we consider adding just one more percent to the efficiency of a glider.  The required lift to drag ratio is now 100, as of yet unachieved, and presently seen as unachievable in unassisted gliding flight.
Regarding propellers, my knowledge is rather limited.  My general discoveries regarding propellers have shown me that there are some pretty good methods for calculating the thrust of a propeller, but that the most accessible and empirically based methods give only a reasonable estimate of propeller thrust and performance.  What is the issue with propellers?
The propeller is usually the last thing considered in aircraft design.  Why is that? When we get a design on paper, we essentially know all of the aspects of thrust and power required for flight.  We get everything down to the powerplant and weight and balance, except for the propeller.  Then we find we need to develop the required thrust by conversion of our powerplant output to real, useful work in propelling our aircraft. Now we need to select a propeller.  We now have to determine if we build our own propeller from scratch, or purchase one already made, in a generic sense, from a reputable manufacturer.  Aircraft design does not start with the propeller first, and then development of an airframe for fitting.  It rather starts from the airframe first, and then deals with the propeller as a special problem. Elsewhere, Peter Kämpf and sophit have given excellent answers for consideration.  Look at what they have said regarding propeller thrust calculations.
There is no desire here to lay out chapter and verse of flight theory in correcting a misunderstanding.  But the following will be said:  Powered flight comes down to three basic elements, namely weight, drag, and velocity.  These elements are opposed in flight by lift, thrust, and power. Each of these elements has a manifested and interconnected relationship with the others, as follows -

static factor
aerodynamic requirement
the relationship

weight
lift
= (power/velocity)/(lift/drag)

drag
thrust
= weight/(lift/drag)

velocity
power
= velocity x thrust

If necessary, we can state these interrelationships in general algebraically defined terms.  Skeptical??  Just write out the equations as equalities in the relations and see what happens as each element is changed in definition.  Here, in basic terms, no equations yet, using the British engineering system, our expression
$$(lift^2/drag)*(velocity)=power$$ becomes $$lb * ft/s = power.$$  OMG!
So here are some thoughts. There are likely some who will read this answer and say it has no place on SE-Aviation.  My leaning is in agreement with that sentiment.  Nevertheless, the OP's questions, and answers, are valued.  The hard part is learning thru developing the answers; teaching thru developing the questions.

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If you're designing a propeller airplane, propeller is definitely not the last thing considered but it's just as important as any other component: number of blades, its size, rpm, with or w/o variabile pitch, its position,... there's a lot to be decided and finally the performance of the whole airplane depends on it.
Power doesn't equal velocity rather it's needed to gain velocity and to maintain it, otherwise it'd go to zero due to friction (aerodynamical or mechanical); if there's no friction (think about space shuttle in space) then there's no need of any power to keep the  velocity.

@RobertdiGiovanni The answer provided by me never made the statement power=velocity.  Further, the propeller consideration process in the answer is a summary based on observed aircraft development during WW-2 and Patey's work, particularly on Scrappy.

@ThomasPerry no one here is saying power = Velocity.  What I have said is velocity cancels out of P/v.  This was my error (the original falling rock analogy) which the writers here have so graciously corrected.

The power consumed by drag in a gliding aircraft is simply its weight times its best rate of descent in feet per second. So a well-designed glider glides on less than 20 horsepower. This means to turn the glider into a plane, it would need an engine of more than 20 horsepower in order to climb.
Similarly, the power needed to climb is the weight of the plane times the desired vertical speed, plus the drag horsepower which can be found by solving for the glide equation above (and assuming the airspeed in your glide is equal to the airspeed in your ascent).
BTW If you tie one end of a tension gauge to the tail of a 172 and chain its other end down and firewall the engine, you will not register anywhere near 900 pounds of thrust...

Most gliders sink at well below 10hp (e.g. an LS-4 glides at less than 4hp) so 20hp seems a bit overpowered :)

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text:
How much does water injection improve thrust of a jet engine?
In 1961 a Navy F4 Phantom Skyburner broke the world speed record with the aid of a water injection system to boost thrust.  Water injection was also commonly used to improve take-off performance of many aircraft of that era.
The application in question would be a steam thrust mechanism for a hypersonic propulsion system.
Modern jet engines compress air, raising its temperature to around 1000 K, before heating it to around 2000 K from combustion of fuel.
This would be a pressure or volume gain of around 2 to 1.
Water to steam results in a pressure or volume gain of 1600 to 1.
Hypersonic aircraft surfaces routinely are heated by friction to over 2000 K.
Question 1: why are they even bothering to burn fuel when plenty of heat energy is available from aerodynamic friction?
Question 2: how much thrust would be generated from water to steam vs combustion/heating at those speeds?
That's where the issue is.  Essentially, the steam motor would be a rocket.  Mass flow would be much less but thermal expansion, 1600 vs 2!
The Rolls-Royce Olympus 101 early jet engine generated around 11,000 lbs of thrust, 9300 lbs from momentum thrust and 1800 lbs from pressure thrust.  What sort of numbers would the "steam engine" generate in comparison$^1$?
$^1$ peroxide as monopropellant may be used as a reference. (Temperature of decomposition into steam around 1250 K).

On the Harrier water can be injected into the engine during hover to keep it from overheating. Run out of water? Time to land, or at least to stop hovering.

Tell us about this routine hypersonic aircraft please.

@OrganicMarble hypersonic vertical flight is indeed routine since the 1940s.  At issue here is the viability of using external heat to generate an expansion reaction in sustained level flight.  Keep in mind 2H2 + O2 ---> 2H20 combustion (around 2700 K) may not be needed, simply carry water.  At 120000 feet even hypersonic flight generates very little dynamic pressure (hence reaction thrusters).  I was hoping (maybe tomorrow morning) someone could flesh out the thrust and water consumption numbers.

Still waiting to hear about routine hypersonic aircraft. Any example will do.

@OrganicMarble this question is about a propulsion system, not an aircraft.  But, if you would like, something similar to a DreamChaser.

Just wondering why you said this "Hypersonic aircraft surfaces routinely are heated by friction to over 2000 K." When there aren't any hypersonic aircraft flying routinely.

@OrganicMarble  OK,  when they do fly (still not very often, working on it),  high heat is routine.  Happy New Year.

When used 60 years ago on the B-52 and 707, water injection increased thrust by about 30%.  Enough water was carried on the 707 for about 5 minutes of takeoff and climb.  Its system and the water added weight that could no longer be justified when engines became more powerful.  The Concorde was slightly underpowered in dry thrust to be able to operate on 12,000' runways and this was addressed by adding about 20% more afterburning thrust.  Considering these examples, the effect of water injection is not insignificant, but it seems it is no longer ever the preferred option.  Besides system weight, another drawback is that it is not environmentally friendly and due to the cooling, not all fuel is combusted, leaving soot and pollutants.
Note that water injection increases thrust by accelerating the water mass (and also increases engine performance by cooling); similarly, the mass of added fuel in afterburning also accounts for part of the added wet thrust.  In this sense, given that afterburning engines get around two thirds more thrust, effectively by quadrupling fuel consumption, when a significant portion of that thrust is from accelerating the fuel mass, not just from combustion, it is understandable that we might start to wonder if we should swap a few hundred gallons of fuel for water.
I think the numbers on SFC (specific fuel consumption) for afterburing engines are worth commenting on here because the inefficiency is insane, whereas water is rather cheap.  For afterburning turbofan engines, fuel flow usually quadruples in reheat.  The SFC in the F-16's FW100 increases from 0.76 dry to 1.94 wet - dry thrust being 17,800 lbs and 29,160 lbs wet.  When calculating with FW100 thrust figures, dry fuel burn at sea level is 225 lb/minute, but in reheat/wet it's 943 lb/min.  So by more than quadrupling the fuel burn, the result is just 64% more or 11360 lbs additional thrust.  Considering the F-16 has 7000 lbs internal fuel, this is a heavy price to pay for 64% more (sea level) thrust and it certainly makes an additional 30% thrust from water injection sound appealing at first glance.
As far as actually commercially "budgeting" for wet thrust, the Concorde was the only afterburning commercial aircraft, so it would be interesting to know how close they were to adopting water injection.  It would appear the weight savings with afterburners was more important than the cost savings of a heavier water injection system.
For the purpose of civil experimental aircraft it would be extremely interesting to see how water injection could be employed and if a similar 30% boost is achievable.  Given that the small TJ-150 engine produces 337 lbs max thrust and probably burns 50 GPH at sea level, the prospect of getting an additional 100 lbs thrust for minimal cost or weight, even for a few minutes, would be very appealing.
Edit: I should also add that the Honeywell TPE331 turboprop engine gains an additional 10% power - Augmented 5 min Rating - from water methanol injection.

Water to steam would have a specific impulse of 100-200, but it's cooling effect probably allows for increased fuel without overheating (hence "sooty" exhaust).  It's any thrust at very high altitudes I'm after, using the atmospheric friction from hypersonic speeds as a heat source. Like the TJ-150 application though. +1.

With pure jet engines the air density is a major factor in thrust. By injecting water the density is increased. I worked on both the B52G and KC135A military aircraft using J57 engines. Both had water injection which was used on takeoff. Loss of water injection during take off was a emergency that pilots feared. Each engine had a regulator valve to control the amount of water going to that engine. A defective valve would cause the that engine to use far more water than normal resulting in the water supply being exhausted during the take off roll or shortly after the wheels leaving the runway.
Later versions of both aircraft used turbofan engines which did not use water injection.

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Is $\eta \frac{P}{V} $ using aircraft speed valid for thrust calculations?
A writer recently used Thrust = $\eta \frac {P}{V}$ here to determine the maximum forward airspeed of an aircraft for a given set of conditions.
While efforts made to answer were very impressive and reflected a lot of effort, there was a nagging question that seems difficult to reconcile:

does thrust always decrease when airspeed increases?

Wouldn't a better expression of prop thrust be:

RPM × AoA$^1$ to the relative wind?

Before the mass flow argument comes in, let's consider that specific impulse of props are in the thousands while the best (mass flow) rockets barely top 400. Props pull by making lift, and loss of $\Delta$ mass flow velocity will have far less effect on thrust than non-optimal AoA at higher airspeeds.
The drag of the prop is consuming the engine power, not the planes forward speed!
Which leads us to:

jets are constant thrust, props are constant power?

Really?  Isn't engine power strictly a function of fuel flow?
As an aircraft gains airspeed, prop pitch must coarsen to maintain optimal AoA.  Thrust loss is only cosine of this angle, in other words, not much at all at lower angles, especially if prop RPM is high, which brings us to this, courtesy of Stipa and Caproni from the 1930s.
I know, too slow, not enough power, too much drag.  But if the propeller were recessed into the duct where air was slower and higher pressure, might this not be an analog of today's fan jets?  Could ram pressure at increasing airspeeds eliminate the need for variable pitch?
$^1$ as part of a prop "lift equation"

The same author has also linked this answer which better explains how that equation is derived an its limitations. If it's anyway not clear maybe you can ask to expand that answer.

(Prop rpm)$^2$ × prop lift equation at optimal AoA × cosine deviation angle from thrust line (coarsening to maintain AoA at higher speeds) × a $\Delta$ Mass flow factor (small but significant)  seems a bit better than a fudgy $\eta$.  But the "authors" number seem OK.  Just curious.  Power output of the engine seems to get confused with the kinetic energy state of the moving aircraft F×V or mV$^2$/s.  Engine power only turns the prop.

Well, if a simple $\eta$ is enough, there's no need to overcomplicate it  Also because the next less-simple mathematical model to represent a propeller would be the Blade Element Theory that, albeit simple in its hypothesis, is quite lengthy in its results.

The drag of the prop is consuming the engine power, not the planes forward speed!

Power output of the engine seems to get confused with the kinetic energy state of the moving aircraft F×V or mV2/s. Engine power only turns the prop.

There's a bit of misunderstanding here.
Most, if not all, of the engines just make something rotate: a car's engine makes its wheels rotate; a drill's engine makes a drill bit rotate; a propeller's engine makes the propeller rotate and so on.
But these rotations are not connected to any power until they create useful work. An engine of a car which is on a perfectly iced surface is not using any power even if it's spinning at its max rpm, except for the power needed to win its own internal mechanical friction. An engine of a drill which is just turning freely without doing any hole is not using any power except the one needed to win its own internal electro-mechanical friction. An engine of an airplane with a propeller that is not producing any thrust, is not using any power if not the one needed to win its own internal aero-mechanical friction.
Only when these rotations create a work then there's a power associated to them. The car's engine produces power when the wheels make the car travel at some speed $V$. The drill's engine makes the electric meter spin very fast only if it's drilling a hole in a reinforced concrete wall. The propeller draws power only when it pulls the airplane at a speed $V$.
So, connecting airplane's speed with propeller's thrust and with engine's power does actually make sense and it's physically correct. And since we do not live in a perfect universe, some power get lost in the process and we model this lost via a simple efficiency factor $\eta$.

Wouldn't a better expression of prop thrust be:
RPM × AoA to the relative wind?

Indeed, $\eta$ depends on the blade pitch (AoA) as measured at ¾ of the span and on the advance ratio $J=\frac{V}{nd}$, where $n$ is the rotating speed [rev/s].

By including the efficiency factor $\eta$, the author gave the expression $T=\eta\cdot\frac{P}{v}$ the needed flexibility to be correct for a propeller and piston engine driven airplane. Efficiency is not constant, and especially at high speed might suffer from too small angles of attack (for fixed pitch propellers) or Mach effects (for all propellers).
Just look at the propeller efficiency for different pitch settings, plotted over advance ratio J (below).

Thanks Peter.  Does it not seem the "efficiency" is pitch related?  Does an aircraft doubling airspeed always mean half thrust for the same engine power output?  (Not to be confused with the aircraft "power state" which is Force × Velocity).

@RobertDiGiovanni Yes if efficiency does not change. Since engine power is constant and power is force times speed by definition, the only reason why thrust won't be half at twice the speed is a change in efficiency.

the propeller draws power only when it pulls the airplane at a speed V

Let's look not at thrust loss at higher speed, but on the runway at full throttle with brakes set.
According to the formula Thrust = $\eta\frac{P}{V}$ when the brakes are released thrust is nearly infinite.  Is this actually true?
Another Cessna 172 rolls onto the runway at 3 m/s (around 6 knots), which one gets off the ground in the shortest runway distance?
Both accelerate at 3 m/s$^2$.  Take off speed is 24 m/s.  Easy one, rotate after around 8 seconds.  But the rolling start bought me 1 second of acceleration.  The last second speed increases from 21 m/s to 24 m/s, averaging 22.5 m/s.
22.5 meters, or about 74 feet shorter takeoff run.
Will the plane sitting at full throttle have infinite thrust when brakes are released?  Will acceleration markedly drop as it rolls forward?
It seems, as theoretically and mathematically correct $\eta\frac{P}{V}$ is, the proof would be to watch takeoff rolls of jets and props.  Increasing drag should begin to reduce acceleration, but at low speeds force = mass × acceleration, at higher speeds acceleration is 0 (steady state) and force = drag.
It may be the prop acceleration curve from start to rotation is straighter than we think, even for fixed pitch aircraft$^1$.
$^1$ increasing efficiency may actually increase rate of acceleration during takeoff roll.

"According to the formula when the brakes are released thrust is nearly infinite. Is this actually true?"
No it's not and I have already pointed you to this other answer of mine where I explain its limitations and what to use instead.

Oh, BTW HP force (could) = mg weight + m$\Delta$V/t weight + m$\Delta$V/t Horse (+ Drag Horse and weight + pulley friction).  Last 2 are minimal.  No mechanical advantage single pulley, though I did consider adding a gear for one poney (or having 2 ponies).

But once the Horse got going (to steady state) HP required = mg weight (+ Drag Horse and weight + pulley friction).  I got a real kick out of comparing these thrust formulae to the thrust curves of Estes model rocket engines.  Big spike to get started, then a plateau until burnout.  Back in the 1700's, oats and hay were more easily had than rocket fuel ;-).

"Some of the formulas tried didn't scale well." which one? I'd be glad to explain (or correct) it better. Have you double checked the units? And btw what's your design?

@sophit I'll put it in a new question.  But try this one:  footpounds of torque (at a given rpm) ÷ feet 3/4 prop arc radius = pounds drag. Pounds drag × 4 ($\eta$ factor) = pounds thrust.  Way less L/D than glider wing but props are in the 300-400 mph range.  (I'll go metric eventually). (Should work for 2 or 3 bladed props).  Good day.

Metric yes please

Let us continue this discussion in chat.

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text:
What was the Vbg and glide ratio of the Rutan VariEze and the Beechcraft Starship?
These pusher designs were marketed featuring high cruise efficiency.
No doubt the pusher configurations offered advantages in propulsion efficiency and cabin comfort, but how did this compare to aerodynamic efficiency?
What was the Vbg and glide ratios of the Rutan VariEze and Beechcraft Starship aircraft as compared with, for example, a Beechcraft Model 18 or a Piper Cheyenne?

Below is the glide distance diagram out of the Varieze POH.
As I read this, it looks to be about 10:1 with engine out and a windmilling prop.
The Long EZ is a tad better:

It looks to be about 14:1 with engine out and windmilling prop. In idle it is about 17:1. That works out to be about a 3.2 degree glide slope, only slightly steeper than the standard 3 degrees.
I fly a Long EZ, and I can tell you from experience, one of the hardest things (not that anything is hard, it's a joy to fly), is to control airspeed in the traffic pattern, the airframe just wants to go fast. In idle power, on a normal glide slope, unless you have a stiff headwind, you almost always have to either slip it a bit, deploy the landing brake or fly a lower, shallower final to keep the speed down.
For all canards, this is because, unlike conventional tailed aircraft, where the horizontal stabilizer must be generating a downwards aerodynamic force in order to create the pitch up moment necessary to compensate for the pitch down moment of the lift on the wing, the canard, being at the front of the airframe, is also generating an upwards aerodynamic force to do that. So the total aerodynamic force, (and corresponding trim drag) is less.

Good data to start.  Improvements with the gear up Long, and where these really shine is in cruise.  Yes, I think I would use that speed brake and come in a little steeper.  Need info on the Starship, these first 2 are holding their own.

@Robert, Exactly! In case of engine failure on final! Nice to be able to just raise the speed brake and have extra glide distance available. Also, for need to go-around.  In T-37's in UPT, we used speed brakes on final so we could carry higher rpm for go-arounds because of long spool-up time required on centrifugal flow turbojets.  And sorry, don't know about Starship...

I fly gliders, but not a LongEZ. However, I believe the reason why the LongEZ is efficient is the same as the glider, and has nothing to do with the canard vs empennage. It is due to the very narrow fuselage profile, attained by having the pilots be practically prone. There's a great site on this, created by someone who was developing a canard homebuilt: https://www.apollocanard.com/4_canard%20myths.htm

Well, I guess any and all performance metrics are affected by all airframe characteristics, no single characteristic can be held to be accountable.  But given the same other general characteristics, the greater the drag, the worse the glide ratio will be. And an airframe with a canard has lower trim drag than the same airframe would have with a conventional tail.

And, in reading the referenced article, it has some clear obvious mistakes. It mentions, several times, that canards limit AOA, and therefore cannot reach their maximum lift coefficient. This is irelevant. Minimum drag (L/Dmax),  occurs at significantly lower AOA than at the maximum or stall AOA.


text:
Thrust = Drag. What about Power?
Understanding all this has been a work in progress, but now from start to finish we have:
Fuel/second ---> shaft Horsepower---> RPM ---> Thrust ---> Drag

Drag Force = $\rho$ × Area × Cdrag × V$^2$

Velocity is proportional to $\sqrt{Drag}$.
Transmission efficiency of a properly pitched  propeller can be considered (as 1-sin $\theta$), or around 0.8.  In other words transmission losses are only slightly worse than automobiles, which have direct force transmission to the ground.
All well and good.  Power, as % BHP in the POH, is the throttle, or fuel flow rate.  RPM is proportional to the square of fuel flow rate.  But the propeller "steals back" efficiency (by V$^2$ in its equation) so thrust is proportional to fuel flow rate. Velocity is derived from Thrust = Drag as $\sqrt{Drag}$.
Another, much simpler example:  parachute and weight.
Double weight and velocity increases by the $\sqrt{}$ of increase of force, or 41%  No need to go up to "Power terms".  Thrust = Drag!
So, what about multiplying Thrust or Drag by velocity to compare with another Force × Velocity.  For example, we know that Drag will be 9× more at 3× speed.  Yet if we use, for example, 120 kg of prop drag at 450 km/hr, is that the same or comparable to 40 kg of airplane Drag at 150 km/hr?
450 × 40 = 150 × 120 = 1 : 1.     No
Only when velocity is squared and the correct Drag value is plugged in does the equation work as:

V1$^2$ × Drag 2 = V2$^2$ × Drag 1

The correct value for Drag at 150 km/hour is 13.3.
Not sure about the value of converting Thrust or Drag into "Power" when "steady state" seems to be a balance of forces at the same Velocity.
Data for this question is taken from the Cessna 172 POH as follows at 2500 feet:
RPM....% BHP.....True Airspeed....GPH.....Range
2700.......87 %.............139...............9.6.......545..
2600.......78 %.............133...............8.6.......590..
2500.......70 %.............128...............7.7.......630..
2400.......63 %.............122...............7.1.......655..
2300.......57 %.............116...............6.6.......665..
2200.......51 %.............109...............6.2.......665..
In question is the relationship between GPH and RPM, which suggests that fuel is converted into piston Force (and Heat), and $\Delta$ RPM is to the square$^1$ of that Force.  So is % BHP!  And airspeed.  Mechanisticly, this suggests as follows:
Fuel potential energy --> Force/Heat --> Force$^2$ = Prop $\Delta$V
Prop "lift" equation $\Delta$V squared --> Thrust = Drag
 Resultant Velocity proportional to $\sqrt{Drag}$.  The chart data seems to support this.
$^1$ not cubed

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"Velocity is $\sqrt{Drag}$" is not a meaningful statement.  It confuses causality and is simply wrong.
Drag (at fixed drag coefficient) is proportional to Velocity squared.  $D=0.5 \rho\, C_D\, V^2\, S$.  However, when you change velocity, you must also change your lift coefficient -- which changes the drag coefficient.  The parasite term of drag is proportional to $V^2$, but the induced term ends up proportional to $1/V^2$.
At cruse, when thrust is equal to drag, we sometimes call the power associated with thrust the "Thrust power".  It is equal to the product of thrust and velocity.  Full stop.  $P=T\, V$.  Since drag is equal to thrust, we can also think of this as the "Drag Power" $P=D\, V$.  It can be thought of as the rate of energy dissipated into the air due to drag.
If you want the power associated with flight, this is it.  You're done here.  However, you also want to see the connection from here through to fuel flow.
The propeller efficiency (say $\eta_P=0.8$) means that the shaft power is larger than the thrust power $P_{shaft}=P/\eta_P$.
The energy contained in the fuel is characterized by the lower heating value ($\mathrm{LHV}$).  We think about fuel consumption for a piston engine in terms of brake specific fuel consumption ($\mathrm{BSFC}$).
BSFC is commonly used in the terrible units of lbm/(hp⋅h).  I.e. pounds of fuel consumed per shaft horsepower each hour.  So, the fuel flow rate looks something like:
$\dot{m}_f= \mathrm{BSFC}\, P_{shaft}$
The $\mathrm{BSFC}$ is related to the $\mathrm{LHV}$ by the cycle efficiency of the engine $\eta_c$.  Obviously a combustion engine is not perfectly efficient -- it gives off substantial waste heat to the air (whether air cooled or via the radiator if liquid cooled) as well as waste heat in the exhaust.  All that waste heat is lost energy that was originally contained in the fuel.  The cycle efficiency of a piston engine is very low -- typically less than $\eta_c < 0.3$.
$BSFC=\frac{1}{\eta_c LHV}$
We can combine these equations together...
$\dot{m}_f= \frac{D\, V}{\eta_P \eta_c LHV}$
Or, another step further...
$\dot{m}_f= \frac{0.5 \rho\, C_D\, V^3\, S}{\eta_P \eta_c LHV}$
If we assume a simple parabolic drag polar $C_D=C_{D,0}+K\,{C_L}^2$, then we get...
$\dot{m}_f= \left(0.5 \rho\, C_{D,0}\, V^3\, S + \frac{K\ W^2}{0.5 \rho\, V\, S}\right) \frac{1}{\eta_P \eta_c LHV}$
This gives a parasite power term that is proportional to $V^3$ and an induced power term that is proportional to $1/V$.
IMPORTANTLY, this is not the whole story.  Buried in this are the facts that propeller and cycle efficiency are not constant.  Propeller efficiency $\eta_P$ will vary with speed and throttle setting (power).  Cycle efficiency $\eta_c$ will vary with throttle setting (power).  In particular, most engines are most efficient operating at near maximum power -- and are very inefficient operating near idle.
If you look at a given aircraft's POH and consider the fuel flow values, you will not see perfectly $V^3$ behavior because of many things -- there is also the $1/V$ term -- and the variation of $\eta_P \eta_c$ is complex and nonlinear.
Edit Below
If we divide the fuel flow by $V$, we get the specific range -- a fuel economy measure like MPG for a car.
$SR= \left(0.5 \rho\, C_{D,0}\, V^2\, S + \frac{K\ W^2}{0.5 \rho\, V^2\, S}\right) \frac{1}{\eta_P \eta_c LHV}$
Which will work out in ft/slug or m/kg, but can be thought of (converted to) nmi/lbm or mpg.
You repeatedly assert that observation of POH data leads you to conclude that various behavior is linear.  This is an erroneous conclusion.  Taylor's series for any smooth function is:
$f(x_0+\Delta x)=f(x_0)+f'(x_0)\Delta x+\frac{f''(x_0)}{2}(\Delta x)^2+\frac{f'''(x_0)}{6}(\Delta x)^3+...$
When $\Delta x$ is small, then $(\Delta x)^2$ and other higher powers will be smaller and smaller such that they can be ignored.  This leaves us with a linear function:
$f(x_0+\Delta x)=f(x_0)+f'(x_0)\Delta x+...$
I.e. everything is linear on a small enough scale.
And on a slightly larger scale, you can keep the quadratic terms and drop everything higher...
$f(x_0+\Delta x)=f(x_0)+f'(x_0)\Delta x+\frac{f''(x_0)}{2}(\Delta x)^2+...$
I.e. everything is quadratic on a slightly larger scale.
If you are looking at POH data, the manufacturer will supply data points at a resolution such that a pilot can linearly interpolate between any given points (or possibly just use the nearest point) to get a reasonable answer.  As you point out, we know that fuel economy is bad at low speed -- so the POH does not give data there.  It is not useful to a pilot.
However, if we are looking to understand the physics of flight, then we prefer to look at a wider range of data.  A range sufficient to show the higher order terms and the more complex interactions.
Edit 2 Below:
The claim was made that thrust can not increase faster than V^2.  Below is a plot of thrust available and thrust required (drag) for an F-5E at 36,089 ft altitude.

If this is insufficient to show that drag can increase faster than $V^2$, we can overlay some analysis...
Here I take the lowest drag point -- since this line corresponds to L=W, it is the point of best L/D.  If the drag polar were of simple parabolic form, induced and parasite drag would be equal there.  I then plot lines proportional to $1/V^2$ and $V^2$ that go through the half-minimum-drag point.  I also plot their sum.

The gold line does increase slower than parasite drag's $V^2$ -- because of the $1/V^2$ term of induced drag.
We also see that neither simple curve keeps up with the real drag.  This is because of the changes in $C_{D,0}$ and $K$ with Mach number and the non-parabolic parts of the drag polar at high $C_L$ (for the low velocity region of the curve approaching stall).
In the real world, these non-ideal behaviors can drown out the idealized trends.  In this case, in the transonic region, $C_{D,0}$ more than doubles between M 0.9 and 1.1 and then remains nearly constant supersonically.
For a piston prop, you won't experience this -- and admittedly, it is one of the most drastic changes in the physical world.  They called it the sound barrier for a reason.  Unfortunately, I don't immediately have my hands on a similar original chart for a piston-prop aircraft.
Note that the textbook approximations for jet aircraft performance would also tell you that thrust available from a jet is constant with Mach number.  It clearly is not.

$\eta$ is not that complex approaching it from the car point of view.  Up at cruise AoA is changing little. Prop efficiency is near 1.  But you are absolutely right when things get slow, AoA increases, lift vector tilts back, airflow becomes more turbulent.  That's why it is fascinating that Vmin power is at the slow end.

I wasn't implying drastic behavior in $\eta_P\eta_c$.  I was implying behavior that is not easily put into an analytical form.  I should have mentioned a few more, $C_{D,0}$ will change with Mach and Reynolds number, and $K$ will change with Mach number.  The common textbook assertion (for example) that best range is at best $C_L/C_D$ for a piston prop aircraft is based on constant $\mathrm{BSFC}$, $C_{D,0}$, $K$, $\eta_P$ -- none of which are true.  Most books do not make it clear that these are approximations -- not gospel.

So, as mentioned with Aditya, steady state is a maintenance of T = D.  Power term V falls out because V is equal, in this case T and D are on the same prop.  If one changes V, yes, additional force is required to overcome inertia, but the drag at V2 =Thrust at V2.  But drag and thrust are not linear to V.  My search has yielded that thrust must increase by the square of Velocity increase, but not more (theoretically).  However factors you mentioned such as AoA, Mach numbers, etc. certainly play a role.

T=D is a statement about conservation of momentum.  Thrust power equal to Drag power is a statement about conservation of energy.  There are certainly times when thrust can increase faster than V^2.  I'll post an image as another update.

First of all, velocity is not $\small \sqrt{drag}$  ; velocity is proportional to $\small \sqrt{drag}$. Additionally, as Rob McDonald said, this is true only for a constant drag coefficient, which is not the case in practice. And as Sophit said, at low speeds, total drag reduces with increase in speed. This is because at those speeds, the effect of reducing $C_{di}$ is greater than that of increasing dynamic pressure.

Fuel/second ---> shaft Horsepower---> RPM ---> Thrust ---> Drag

I would disagree. You start with an expression for Power (fuel/second), proceed with an expression for Speed (RPM), and finally end with an expression for Force (drag).
I my opinion, it would be more appropriate to end with the quality that we started with. If we start with power, we must end with power. The order should look like:

fuel power ---> shaft power ---> thrust power ---> rate of doing work

$fuel\ power = \large \frac{fuel\ energy}{time}$
$shaft\ power = rotation\ speed \cdot torque$
$thrust\ power = speed \cdot thrust$
Rate of doing work
The thrust power does work on:

The aircraft

The air particles

Energy does work, and so rate of doing work is same as rate of change of energy. So let's look at rate of change of energy of aircraft and air particles.
Aircraft: It possesses energy in two forms, kinetic and gravitational potential. Therefore, the rate of change of energy of the aircraft is:
$$\left( ½ \cdot m \cdot \frac{dv²}{dt} \right) + \left( m \cdot g \cdot \frac{dh}{dt} \right) $$
Air: The aircraft does work on air particles by imparting kinetic energy into them, and this manifests itself as drag. The work done on air is:
$$½ \cdot ρ \cdot V³ \cdot C_d \cdot S$$
this is known as drag power $-$ the power required to overcome drag.
Therefore, the total work done by thrust power (in a unit time) is:
$$thrust\ power = \left( ½ \cdot m \cdot \frac{dv²}{dt} \right)  +  \left( m \cdot g \cdot \frac{dh}{dt} \right)  +  \left( ½ \cdot ρ \cdot V³ \cdot C_d \cdot S \right) $$
Note here that if thrust power is less than drag power, then either $\frac{dv²}{dt}$ or $\frac{dh}{dt}$ (or both) will be less than zero. This means that the aircraft will either lose gravitational potential energy, kinetic energy or both. Likewise, if thrust power is greater than drag power, $\frac{dv²}{dt}$ or $\frac{dh}{dt}$ (or both) will be greater than zero.

is kinetic energy ever "transfered" in propulsion?  Or is momentum transfered?.  With drag increasing V "strikes twice as many molecules twice as hard", which is 2 × 2mv.

@RobertDiGiovanni Both KE and momentum are imparted. Assuming constant $C_d$, If speed increases by a factor of 2, momentum imparted will increase by a factor of 4 and kinetic energy by a factor of 9.

I think we are essentially in agreement with thrust = 1/2m dv$^2$/dt. Prop RPM is proportional to fuel burn rate$^2$ because the cylinder Force as Average psi/cylinder Area must work to equal prop drag.  Essentially, fuel potential energy is "downrated" to mv in propulsion of the pistons.  Therefor it takes the square of fuel flow to increase RPM (see POH).  But thrust from the prop is a function of V$^2$.  I'm fairly sure that this is why early jets used a lot more fuel than fan jets or turboprops. (afterburners too).  The propeller greatly improves efficiency (for a Cessna 172 speed range).

@RobertDiGiovanni: $\left( ½ \cdot m \cdot \frac{dv²}{dt} \right)$ is the time derivative of KE. KE is energy, and its time derivative $ \left( \frac{d(energy)}{dt} \right) $ is power.  In the LHS, you have thrust, where you should instead have power (so thrust must be multiplied with speed). Also, this is true only when $C_d$ (drag) is zero and $∆h$ (instantaneous change in height) is also zero.

And the reason RPM is proportional to (Fuel Flow)² instead of (Fuel Flow)³ might be that the engine is quite inefficient at low RPMs. If the prop was driven by a constant efficiency engine, I'd expect RPM to be proportional to (Fuel Flow)³.

I believe the mathematics is correct.  I only suggest that, in the analysis, based on POH data and the observation that pistons produce force (with heat and friction losses) with a certain fuel consumption at a certain RPM.  We are left to determine whether we pay drag (for 2x speed increase for example) with 2 × 2mv or some sort of Power term.  Both ways of explaining work.  Interestingly, we pay for drag twice, once for the prop and once for the airplane!  But prop efficiency makes fuel/time 4x for 2x speed.  Fuel consumption is proportional to V for distance

I'm sure we'll have more conversation about this in the future, and appreciate your writings.  Robert

@RobertDiGiovanni If the propeller speed is doubled, power required to drive the propeller should increase by a factor of 8, and the following is why I think so: the drag (torque) experienced by the blade increases by a factor of 4 (since drag is proportional to the square of speed). But also, the crankshaft is rotating at twice the RPM, so when this is multiplied by 4 times the torque, we get 8 times the power. Maybe, the increasing engine efficiency is what's making the difference... (note: there's a mistake in my first comment, I meant to say "factor of 8", not "9")

that's why I studied the POH so carefully.  Perhaps the answer lies in the distinction between inertial physics and steady state.  The only thing trying to slow rotation is drag force.  If more fuel is added, piston force > Drag. RPM increases until Force = Drag.  RPM is essentially how fast the fuel pump is running.  But I will keep my mind open. Alas, the difference between 2 squared and cubed is 1.41 vs 1.26.  POH and car data support V$^2$, but I will keep an eye on future work.

Let us continue this discussion in chat.

This answer is provided in response to concerns expressed by the moderator regarding extended comments. Elements of the question are very interesting, although the general thread of the question is difficult to follow.  The answer Mr DiGiovanni desires is one that apparently shows the relationship between engine torque (or propeller thrust) and various parameters regarding engine performance, including fuel flow and thermodynamic processes in combustion.
Based on the information that was provided in the question, a considerable amount of additional information can be developed about your plane. However, to answer your question, the following are necessary –

The question must be stated clearly regarding the subject of the question.  This means clearly abstracting the issue at hand, and stating that issue along with relevant information.  Understandably, that is not always easy, and in some cases may take a considerable amount of time and thought.  Keep in mind, a lot of time is spent in trying to answer your question by thinking through and evaluating what it is you are trying to find out.

In the process of providing an answer, data that are provided are usually validated by the person answering your question.  This is usually done by referencing these data directly to a source that shows the same data, or allows the same data to be developed in context.  However, such is not the case with this question.  A desirable solution to this issue is to provide an image of the source of these data that clearly references the document at hand, or provision of a link where these data can be examined directly.  Consequently, for information provided in your question, there is a concern regarding the validity of the data.  Having credible data with unquestionable validity is helpful.

A considerable amount of information can be developed by the data that were provided.  However, the analysis cannot be validated. A considerable amount of time was taken to evaluate your plane based not only on the provided data, but by cross comparison with similar aircraft in the literature.  Although the results of the analysis are interesting and appear to be along the line of interest posed in your question, little else can be done to afford credibility to the answer because of seeming inconsistencies apparent in data provided with the question.  That does not mean those data were in error, just that those data could not be validated and are in question.  Take a look at a summary analysis, for what its worth.  Perhaps this will be helpful…

Really don't need the preamble to the Gettysburg address here, but the graphs are great.  +1.  I spent a week on this, not being a mathematician.  But the similarities of HP and drag are amazing.  Both have units of weight and velocity.  [A][B] = [A][B].  But the trick was realizing fuel is only converted into Force.  The expression becomes A$^2$ = [A][B].  If you plot HP$^2$, this gives the prop drag curve, or RPM, which by way of thrust, is proportional to the aircraft drag.  Plotting RPM would be great if you have the time.

Special request, large graph.  Please divide out V from HP available and replace HP required with GPH.  I'm ready for the "real" Vx and Vy graph.  PS I think RPMs are going higher as V increases because prop AoA (fixed) will get less and less.  More RPM would help offset decreasing prop efficiency.

@RobertDiGiovanni Hey, I think I found your source.  You are looking at the K model, but for which year??  I might be able to pin down a bunch of interesting info you would like to see, and adjust those graphs as you requested.  Tell me your specific POH source, please....

not sure there will be significant differences in data for K's (if there were, it would be a new letter).  I believe it was 1969-1970. Thanks.

@RobertDiGiovanni Ok... Found the source of your data, and supplemental data, both are for the 172M.  We will see how this goes for an update/revision of the answer that was given, including your requested graphs if possible. These data were validated. Data given in your original reference was for airspeed in mph, not kts.  All were standardized to kts in the revisions...

Some definitions. Work, or mechanical energy, equals force times distance. Mechanical power is force times distance per unit of time, or force times velocity.
Let's assume that propeller power output is indeed proportional to engine power, and that engine power is indeed proportional to fuel flow. Thrust is thus proportional to engine power divided by velocity. The other way round, velocity equals power divided by thrust, only works in steady state.
If thrust exceeds drag, we have excess power, which we can put in extra kinetic energy (speed up) or potential energy (altitude). In cruise, all the fuel energy (conventially known as chemical energy, not as potential energy!) is spent on drag, and kinetic energy remains constant. So there's no conversion from fuel energy to kinetic energy.
Drag force is proportional to velocity squared according to experimental physics. So if drag at 450kmh is 120kgf, it is 120/(450/150)² at 150kmh, indeed 13.33kgf. Power is 450x120=54000kgfkmh and 2000kgfkmh respectively, a whopping factor 27 difference (not 1:1 or 3:1)! That is because power is force times velocity, so drag power loss is proportional to velocity to the third power...
No idea where you got that velocity is the square root of drag.

because the propeller has an efficiency near one, more like cars.  For rockets, we are only using momentum as a propulsive force. Like a recoilless rifle.  That's proportional to V$^3$.  A car with direct contact with the road is V$^2$.  Propeller fuel consumption rate is closer to cars with respect to velocity and use of fuel per distance is linear with velocity.  I'm working back from POH data.  Don't mean to make a fuss over this.

Propeller efficiency is not that near to one.  Fuel per distance is not linear with velocity.  All things are linear if you consider them in a small enough range.  You do not capture the nonlinearity of the world if you only consider small ranges.


text:
What effect does airfoil sweep have on lift production?
In discussion of incorporating sweep into vertical stabilizer designs, it is often said that sweep can create more torque moment around the CG because it moves the center of (stabilizer) pressure further back.
Yet, sweep also introduces a different airflow pattern than a straight wing, resulting in air being "plowed" outwards "underneath" as AoA increases, and air being pulled inwards (toward the root) on the low pressure side.
Interestingly, this actually manifests into a useful lifting vortex as sweep increases more, but the Coefficient of lift on the straight airfoil is greater.  The straight airfoil (without high lift devices) will stall at a lower angle of attack.
All well and good, but does anyone have some hard numbers on lifting and stalling properties as sweep progresses from 10-20-30-45-60 degrees?
Would sweeping a stabilizer back 30 degrees actually result in more longitudinal stability, or would it just have more resistance to abruptly stalling at a higher AoA?

This might be helpful

@sophit see page 6.  We're certainly on the right track.

And more here!

There is some content in "Aerodynamics for Naval Aviators" that might help -- https://www.faa.gov/regulations_policies/handbooks_manuals/aviation/media/00-80t-80.pdf -- see pp 74-77-- also 228-232 and especially the uppermost graph on page 228.  I'm not planning to answer at present; feel free to use this in any answer.

@quietflyer excellent content.  So we wonder what a 120 knot Cessna is doing with a swept fin anyways.  Probably more to smooth crosswind effects, and it does look good.


text:
What is the official FAA definition of the term enroute?
I recently saw a ferry permit which included the following limitation:

Flight in IMC authorized for ENROUTE operations ONLY

Great, IMC is allowed enroute!  But wait, where does the "enroute" portion of the flight start and end?

I did some research and found the following ICAO definition for enroute:

Instrument Flight Rules (IFR): From completion of Initial Climb
  through cruise altitude and completion of controlled descent to the
  Initial Approach Fix (IAF).
Visual Flight Rules (VFR): From completion of Initial Climb through
  cruise and controlled descent to the VFR pattern altitude or 1,000
  feet above runway elevation, whichever comes first.

That seems pretty straightforward, but I was looking for something FAA specific and found this in Chapter two of the FAA's Instrument Procedures Handbook:

The en route phase of flight is defined as that segment of flight from
  the termination point of a departure procedure to the origination
  point of an arrival procedure.

That doesn't match the ICAO definition, but it also isn't from an actual regulation or LOI either.

This leaves arrivals and departures as slightly ambiguous, and they often start high in the flight levels where it can be hard to avoid IMC, even on a nice VFR day at the airport (and also seems contrary to what they are trying to accomplish).  Are arrivals and departures considered enroute or some other phase of flight?

According to the information that @aeroalias found - which I'm sure is correct - the ICAO definition is used "internally" by the FAA:

FAA has officially adopted the [...] phase of flight taxonomies
  as an agency-wide data standard for its internal systems.

However, as far as I can see, that hasn't resulted in a clear definition in the regulations or the FAA's publications. There's nothing in 14 CFR 1.1 or the PCG and the best I could find - apart from the definition you quoted from the IPH - are sources that imply that en route (or "enroute") refers to cruise flight only and doesn't include departures and approaches. That would conflict with the ICAO definition.
First, section 2 of the IPH says what en route airspace is:

The en route airspace structure of the National Airspace System (NAS)
  consists of three strata. The first stratum low altitude airways in
  the United States [...] are called Victor Airways. [...] The second
  stratum high altitude airways [...] are called Jet Routes. [...] The
  third stratum allows random operations above flight level (FL) 450.

That's a fairly specific definition that strongly implies that - at least historically, i.e. pre-GPS - en route meant "on the airways".
Second, TERPS doesn't define en route either, but it does repeatedly suggest that approaches and even feeder routes are not en route.
Section 3 on initial approaches says:

In the initial approach, the aircraft has departed the en route phase
  of flight and is maneuvering to enter an intermediate segment

Section 2 on feeder routes says:

[...] the angle of intersection between the feeder route course and the en
  route structure must not exceed 120 degrees

That seems to me to mean that a) by maneuvering to start an approach you are no longer operating en route, and b) feeder routes are not considered en route.
Finally, if you look into some of the FAA's legal interpretations, they seem to say that enroute means cruise flight:
Duncan (2014):

[...] pilots serving as SIC during the en route cruise portion of the flight
  only [...]

Barton (2014):

[...] the pilots identified in both of your questions serve exclusively en
  route, during the cruise portion of the flight, [...]

I assume that most pilots would agree that descending/maneuvering to start an approach is no longer cruise flight. But unfortunately there's no formal definition of "cruise" that I could find - apart from possibly the ICAO definition - so that isn't entirely clear either.
So my take - and I may be completely wrong - is that even in the absence of a formal definition in the regs, there's quite a lot of FAA material that states or implies that en route means cruise flight on an airway or other defined route, and it doesn't include departures and approaches.
But for a 'real' answer, you might have to ask your local FSDO for their opinion, or even ask the FAA for an interpretation if you need it.

You basically redescribed what I found.  ICAO says one thing and the IPH says another, however the examples that you give after that do not say that en route is restricted to cruise, but refer specifically to the cruise portion of the en route flight segment, which actually implies that there is more to it than cruise.  Feeder routes and initial / intermediate segments are all part of an instrument approach, so I agree that those are not en route, but what about arrivals and departures, which is specifically what prompted this question?

@Lnafziger True, but you were asking for additional, authoritative sources so I hoped that what I found might be useful, or at least relevant. Anyway, my basic point is that neither "en route" nor "cruise" seem to have clear definitions in any of the places where you might expect to find them, so there may not be any answer out there unless you ask the FAA directly.

Yeah, this just kind of confirms what I found, although the letters of interpretation that you quote imply that there is more to en route than just cruise (otherwise why would they put the "cruise" qualifier in their statements).  Someone is probably going to have to write the FAA to get a definitive answer.

FAA follows the ICAO definition for describing the phases of flight.
FAA is a member of the CAST/ICAO Common Taxonomy Team that formed the Phase of Flight Definitions and Usage Notes in October 2002. From the CAST/ICAO Common Taxonomy Team site:

Federal Aviation Administration (FAA)- FAA has officially adopted the aircraft make model series (ACFT-MMS) and phase of flight taxonomies as an agency-wide data standard for its internal systems.

So, the enroute phase of flight is the same (just repeating from the same document):

Instrument Flight Rules (IFR): From completion of Initial Climb through cruise altitude and completion of controlled descent to the Initial Approach Fix (IAF).
Visual Flight Rules (VFR): From completion of Initial Climb through cruise and controlled descent to the VFR pattern altitude or 1,000 feet above runway elevation, whichever comes first.
This phase of flight includes the following subphases:
• Climb to Cruise: IFR: From completion of Initial Climb to arrival at initial assigned cruise altitude. VFR: From completion of Initial Climb to initial cruise altitude.
• Cruise: Any level flight segment after arrival at initial cruise altitude until the start of descent to the destination.
• Change of Cruise Level: Any climb or descent during cruise after the initial climb to cruise, but before descent to the destination.
• Descent: IFR: Descent from cruise to either Initial Approach Fix (IAF) or VFR pattern entry. VFR: Descent from cruise to the VFR pattern entry or 1,000 feet above the runway elevation, whichever comes first.
• Holding: Execution of a predetermined maneuver (usually an oval racetrack pattern) which keeps the aircraft within a specified airspace while awaiting further clearance. Descent during holding is also covered in this subphase.

Put simply, en route occurs between terminal procedures.

LAX-JFK
ORCKA1 LAS J146 DVC HBU J10 OBH FOD KG75M DAFLU J70 LVZ LENDY6

Look at the flight plan above, the first and last bit usually indicates the departure terminal procedure (ORCKA1) and the arrival terminal procedure (LENDY6) respectively.  Everything in between is en route.
Another perspective is to think of which controllers own that airspace.  A flight usually starts and ends in terminal airspace and is managed by the local TRACON.  Once leaving TRACON airspace, it is then handed to CENTER controllers along its flight path.  Terminal/enroute boundaries vary by location, but you can think of it generally as Class A airspace

Welcome to aviation.SE! I'm not sure that I understand your comment about class A, there are lots of GA flights that never enter class A but are still handled by Center controllers.

@Pondlife I guess that's a good point.  Ignore my comment on airspace classes.
"The en route phase of flight is defined as that segment of flight from the termination point of a departure procedure to the origination point of an arrival procedure." That's a direct quote from the FAA's Instrument Procedures Handbook.
I guess what's important here is to note that en route is defined in the flight plan which is required by IFR.

Basically, no IFR departures or approaches allowed. Thus reducing the chances of harming the general public during your ferry permitted flight.
After all, a ferry permit implicitly implies there is something amiss with the aircraft, either mechanically or regulatory.

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authors: Марина Лисниченко


text:
How does the mounting location of a jet engine affect aircraft performance?
Different jet airplanes mount the engines in different ways.
For example:

Under the wing

To the fuselage
How does the engine location affect aircraft performance?  Is one better than the other?

There's also the single engine, straight out the tail, like the cirrus sf50

First thing I saw when following that link was seven people digging a hole ;-)

I believe the answers already given are more in line with what you are seeking. However, a minor point to remember is any propulsion out of line with the CG provides asymmetric thrust that must be compensated for aerodynamically. All other factors being equal, power supplied inline with the CG will be the most efficient.

Engine location performance-affecting factors:
1. Mounted in the wing root

low asymetric yaw on engine failure, less rudder required: less drag
no engine pods: less parasitic drag
engines closer to CG, less downforce needed from the tail:  less drag
very little reverse thrust available
Little space for high-bypass-ratio engines

2. Mounted in pods under the wing

high asymetric yaw on engine failure requires larger rudder: drag penalties
engines provide bending relief on the wing, allowing better wing design (thinner wings): less drag
at high incidence angles the pods can prevent spanwise flow: less drag and better stall characteristics
full thrust can impose a large, undesirable, pitch up moment (think stall recovery)
less freedom in roll on cross-wind landings
Location ahead of the wing's elastic line helps to dampen flutter

3. Mounted inside the tail or on pods on the rear fuselage

low asymetric yaw on engine failure, less rudder required: less drag
wing design is freed from the need to accomodate engines, allows for more complex wing designs: better performance throughout the flight envelope
heavy engines so far aft of the fuselage require wings mounted further aft, and a higher tail to support that: more drag
Lower landing gear required, especially in case of short fuselages

That's about what I can come up from memory, maybe someone can merge all answers into a comunity wiki kinda post..

Pictures would be nice. Can you (or someone else) edit in some CC-SA photos or diagrams?

Why does the landing gear need to be lower for rear fuselage engines?

@Radu094 I don't think it's about the landing gear having to be lower, but that the landing gear can be lower due to the fact that it doesn't have to extend past the engines anymore to avoid dragging them over the ground.

What about the HondaJet style mounting above the wing on pods?

Tail mounted are susceptible to less engine damage problems due to FOD I think. Maintenance etc. is easier on the wing mounts. Containment in case of turbine failure is also an issue for close to fuselage designs.

In the part of wing root mounted engines, the point about engines close to CG is invalid. The downforce on the elevator depends on position of wing relative to CG and that can be selected during design, independent of engine position, to achieve desired level of static stability.

What that meant was: in general, wing root mounted will mean less of a fw CG displacement,resulting in an overall rear-wards-CG, all other things being eqal, thus less tail downforce,drag, etc... Though the wording in the answer might be confusing. Feel free to edit it if you want; make it sound better

What about the torque moment created by engines remote from CG and wings? It should it influence the nose down tendency and horizontal stabilizer effect?

You forgot: 1. Mounted in the wing root - High risk of major structural damage if an engine blows up.

Mounting jet engines above the wings (see: Antonov An 72, Boeing YC-14) takes advantage of the Coandă effect to increase lift and improve short-field takeoff and landing performance. The high-velocity engine exhaust (I'm hand-waving a bit here) adheres to the surface and allows the wing to produce lift somewhat longer than it could otherwise.
This layout also vastly reduces the amount of debris and spray that can reach the intakes and damage the engines.

So it's much like using propeller slipstream on a piston/turboprop

Yeah, it's similar in the end result, but I think it's due to a different aerodynamic principle.

Generally there are 3 engine placements:
In the tail, on the fuselage in the back, or under the wings. I'll go over the pros and cons of each.
On the tail
Pros
Not many pros, except that it lets one turn a four-holer into a three-holer which is usually more efficient.
Cons
Generally if you lose the engine in the tail, you risk losing rudder control and you also need a thoroughly reinforced tail, plus you have fuel lines running near the cabin, and you have a significant amount of weight not near the center of gravity.
On the fuselage in the back
Pros
Helps to prevent debris from entering the engine, allows for more room for flaps on the wing, and lighter, more aerodynamic wings since they are not holding up engines, generally allowing for slower approach speeds, also the thrust is closer to the center of the aircraft.
Cons
Fuel lines near the back of the aircraft,tail has to be a T-tail, as well as reinforced. and center of gravity issues.
Under the wings
Pros
Well centered with the center of gravity, allows for more weight shifting in the cabin, and with the bags. The fuel is also kept away from the cabin, and the noise is better spread out through the cabin. In the event that something goes wrong, debris is also less likely to enter the cabin.
Cons
Much easier to get debris swept into the engine, wings have to be reinforced, lose some area for flaps and slats, raising approach speed.
Some useful links
http://adg.stanford.edu/aa241/propulsion/engineplacement.html
http://www.airliners.net/aviation-forums/general_aviation/read.main/750042/
Performance Wise
Tail: Generally not the best, as it has center of gravity issues and drag problems.
Fuselage: Good for poor quality runways or low approach speeds but can add weight in reinforcing the tail.
Under the wing: Good for getting more load as less center of gravity issues, but raises approach speeds.

Not sure your fuel points are really an issue. Many under-wing configurations have fuel tanks in the fuselage right under the passengers, and all of them have fuel cross-feed which means fuel lines are running through the cabin.

You also have fuel lines to the tail for the APU for the underwing configuration.

How do the different configurations affect performance though?

How does a rear engine arrangement require a t-tail? Look at Hawkers, Falcons, Caravelle, etc.

The tail mounted engine also allows single engine jets. Like the cirrus sf50

One con to a fuselage mount setup is that debris from the wings can be ingested into the engines, as with SAS Flight 751. On the flip-side, wing-mounted engines are closer to the ground and more susceptible to ingesting debris from the ground.

HondaJet - over the wing engine mount.  They claim less cabin noise, & I forget what else.

Tail-mounted has many pro's. 2. Back of fuselage - Pro:  "prevent debris from entering the engine" is wrong. Icing from the wing is a big debris problem for this type. 3. Under the wings - Cons " wings have to be reinforced" is wrong. Wings actually have less of a load to support than other configs

I bet the crew of this tail-engined airliner would have given anything to have just lost rudder control.

And there’s the HondaJet. https://en.m.wikipedia.org/wiki/Honda_HA-420_HondaJet

For tail side-mounted: I think there are issues with airflow to engines at certain angles, reducing stall-recovery options.

This is a valid point but this answer could be improved by providing more detail.

I was under the impression that the greater problem was that tail-mounted engines necessitate a T-tail configuration, which is susceptible to entering a deep stall.

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text:
What is a SAAAR / RNP AR instrument approach?
I understand that a SAAAR / RNP AR is a "special" approach and that special approval is required to even get the approach plate, but what exactly is it?  Aspen is an example of an airport that has one.

First allow me to give a quick introduction to RNAV and RNP before geting to SAAAR / RNP AR.
Area Navigation (RNAV) is a method of instrument flight rules (IFR) navigation that allows an aircraft to fly on any desired path within the coverage of referenced navigation beacons, rather than navigating directly to and from the beacons. In other words, waypoints can be defined at arbitrary locations and paths betweens those points (and traditional navigation beacons as well) can be defined.
RNAV routes are coded into the navigation database by ARINC 424 path-terminators. Whilst RNAV routes can be designed in such a way that the lateral path is strictly described, in general the lateral path has a quite some freedom due to floating leg types (e.g. turn at altitude, fly heading etc) and wiggle room for the aircraft to negotiate the turns.
When the route is described only by TF (track to fix) and RF (radius to fix) path terminators, the lateral path is fully fixed. These fixed lateral paths allow procedure designers, for example, to design approaches through mountain valleys down to the runway.
RNAV defines a 95% accuracy bounds. For example, RNAV 1 requires the aircraft to be within 1 NM from the path centreline 95% of the time. The route designer will ensure a 2 NM obstacle free zone around the path in such a case.
Note that RNAV provides no assurance against exceeding the 2 NM zone. There is no automated system monitoring required.
Potentially causes of deviations between the path intended by the procedure designer and the actual flown path can be grouped into three categories:

First, there is the Path Definition Error. This is caused by database
coding mismatch with the original path design.
Second, there is the Path Steering Error, caused by the combination
of display inaccuracies on the Navigation Display and the inability
of the pilot or the autoflight system to stay exactly on path (Flight
Technical Error).
Third, there is Position Estimation Error, which is the difference between the position of the aircraft estimated by the navigation system and the true position.

It is not unthinkable that while normally (95% of the time) the aircraft will fly within e.g. 1 NM of the intended centre line, there are cases that the aircraft will exceed the 1NM by far, possibly even busting the 2 NM obstacle free zone. That could potentially ruin your day if it happens in a mountain valley.
This is where RNP comes in. RNP (Required Navigation Performance) is basically RNAV with added protection. For example, RNP 1 again defines a 1NM accuracy for 95% of the time, but it also offers 99.999% assurance that it will not exceed the 2NM. It will raise an alert if the 2 NM value will potentially be exceeded. So in addition to what RNAV offers, it includes a system monitoring function that will detect navigation errors and alerts for them.
The RNP value alway gives the 95% accuracy, the 99.999% assurance (integrity) limit is always at twice that value. So RNP 0.3 means that the aircraft has to stay within 0.3 NM of the designed path centreline 95% of the time. And if it exceeds the 0.6 NM threshold, an alert will be raised with 99.999% certainty.
RNP systems therefore require a system monitoring function.
To ensure that the RNP values are met, each of the three error sources mentioned earlier must be under control. For RNP values down to 1 NM that is not much of a challenge. Standard procedures and equipment are deemed to be sufficient here. (of course the required RNP values must be met by the Actual Navigation Performance)

If an approach requires tighter limits, such as RNP 0.3 or 0.15, then special authorization is required.
This is where SAAAR comes in (Special Aircraft and Aircrew Authorization Required), now renamed to RNP AR (RNP Authorization Required). RNP AR routes only make use of TF and RF leg types to fully fix the lateral path. RF legs are not supported by all navigation systems by default.
To get authorization for RNP AR, the aircraft needs to comply with the applicable functional and performance requirements, the crew needs to be trained and lot's of documentation must be maintained. Basically proof must be provided that all of the three error terms are under strict control.
In a sense it is somewhat similar to ILS CAT III approval.

References:
AC 90-101A describes a means of compliance for RNP AR
RNAV / RNP presentation gives a good explanation of RNAV and RNP

Does my Garmin 430 have RNP 1.0? Or is it just vanilla RNP?

Hi @JScarry, I don't know what the Garmin 430 supports by default. It should be in the manual. Of course there may be users here that would be able to answer your question if you ask it.

@ymb1 So it looks like WAAS enabled receivers are RNP 1.

@ymb1 This is the kind of thing that makes reading the AIM nearly impossible at times. There are at least four ACs talking about this stuff for different phases of flight and five or more flavors of equipment. I did just find this for the GTN series. It has a software update that allows "RNP 1.0 approval for RF Leg Procedures" http://garmin.blogs.com/aviationalerts/2016/03/gtn-software-version-611-update-and-installation-approval-of-gma-35c.html


text:
Is there any publicly available research that projects pilots' loss of proficiency due to lack of flying (or guidelines to prevent it)?
The FAA requires 3 landings every 90 days to stay 'current', but as we all know, currency and proficiency are different things.
I'm looking for data than can describe how proficiency loss correlates with various factors. Relevant factors might include how long it's been seen you've flown, how much flying you've done in the last ____ days/months, how much time in type, time in seat, or total experience, etc.
And if there are evidence-based guidelines for how much flying a pilot should do to stave off loss of proficiency, I'd like to see that as well.
Thanks.
Edit:
I understand that it's impossible to accurately calculate how proficient someone is based on unique individual traits. That's not what I'm trying to do. Like the title says, I'm trying to quantify what's typical for "loss of proficiency due to lack of flying". In other words, how does flying 10 hours in 90 days compare to flying 10 hours in 6 months, as it pertains to proficiency degradation and, ultimately, increased risk of an accident/incident.

Define proficiency. We use currency because it's something measurable, proficiency is subjective.

"Define proficiency". That's the end goal. I'm trying to gather data and/or research for exactly that purpose.

It's very age/experience dependent, and also dependent on the mental data aspect vs motor skills. A 60 year old that's been flying for years can generally go very long periods and jump in and get right to it, as far as stick and rudder goes.  Those skills stick a long time.  I'm over 60 and don't think twice about going flying my own plane after several months of inactivity. It's like may last flight was yesterday. A young person needs the repetition.  On the other hand, the young person can absorb data at a much higher rate.

@JohnK Understood. It's impossible to determine how proficient someone is with their statistics and demographics alone. That's not what I'm trying to do, and that's not what I asked. I'm trying to determine what the primary factors are that cause proficiency loss, and to what extent they do, generally speaking.
In other words, how does flying 10 hours in 90 days compare to flying 10 hours in 6 months, as it pertains to proficiency degradation and, ultimately, increased risk of an accident/incident.

You should look beyond just aviation for that kind of data.  You'll probably find what you need looking for university studies on that sort of thing for other skillsets that require both mental knowledge and motor skills.  On the military side would a good place to look because it involves the ability to perform and make judgements under stress similar to flying.  Combat skills are drilled to death to make things automatic under stress, just like emergency procedures in aviaton.

@JohnK I honestly do not mean to insult, but age is definitely not a "linear advantage", if any. Just to be clear: age as such means pretty much nothing, as there are far more influential individual characteristics to a person that add up to one's "proficiency" so to say. Age does have the advantage of giving one more time to fly and thus gain experience, but in the end, the ageing body and mind will turn against the pilot. I'm going to be blunt and mention mr. Harrison Ford and his performance as a senior pilot. A senior pilot can be safer than a junior one, but it's all in one's attitude.

Closely related

Well once you're in a mental decline stage, it's another story, but the years between 50 and 70, assuming you still have your faculties, are the best mentally, and motor skills are burned in so effectively they don't decline significantly with lack of use, compared to someone in their 20s.  Ask any 60 year old, and they'll take the 20 year old's body if a switcheroo could be done, but not their brain.

Any decent accident investigation report will show the flight hours flown by the pilot/crew, broken into a matrix of all / acft type in question, and during certain periods of time. These reports are publicly available on different national investigation authorities sites (for example NTSB). The reports also evaluate the currency of the pilot/crew, and mention if it is considered to be a factor in the accident.
If you are looking for an easy way out, in form of a scientific study on these reports, sorry, can't help with that one.
As mentioned in comments, currency in flying does not equal proficiency, and you seem to understand this. Being current does not guarantee one is proficient, not by a long shot, so you are digging a real can of worms here, beware :)
I do think you would benefit reading up on pilot training literature, namely the "human performance & limitation" module. IIRC it contains some very relevant information on the subject of your interest. Such as how experience needs support of other positive personality traits to manifest itself as safety, and perhaps proficiency if you will.


text:
I analyzed 2 nearly identical flying wing in XFLR5, how come the one with the thinner tip chord has a higher cD?
I made two flying wings, one with half the wing tip chord as the root chord and one with the root and tip chord being equal.
I then did VLM2 analysis on both at 7 m/s, and used the same airfoil on both. surprisingly, the wing with the thinner wingtip chord had a higher coefficient of drag (which i believe leads to higher drag). How come?
Everything I know leads to less drag, less induced drag because a thinner chord, thinner airfoil due to proportions, and same parasitic drag because everything else is constant. What causes the one with a thinner wingtip to have a higher CD? Does this make sense, or is it just the inaccuracies of trying to calculate CL and CD?

XFLR5 project file
I designed another plane with the same 30 deg sweep, same airfoil and an aspect ratio of 10 (span .5m, root & tip chord .05 m, no twist). Performance keeps getting worse. Only AOA at 0 makes sense to me but that's out the window.

Are you certain they've been non-dimensionalized properly? Have you tried comparing CL^2 vs CD between the two configs?

I can't find cl^2 so I compared Cl^3/2 vs cd (not sure if this is what you were going for), and it look similar, but again the thinner chord tapers off after 2-3 deg AoA. I ran the analysis through 3d panel and LLT analyses and got very similar results.

What do you mean you can't find it? You just have to plot CL, multiply by itself and plot against CD on Excel

Oh duh. completely forgot you could export data for a second. Yeah gotcha, just did it and updated the main post. Cl^2/cd ratio higher now on the main wing.

Could you share the XFLR5 project file?

Added it onto the originial post.

can i please get the updated project file. Thank you.

In addition to Chris' answer, I have had a look at your project file and could not readily reproduce your problem. I suspect the issue was in a combination of poor discretization (I increased the resolution of the airfoil and wing somewhat) and incorrect methodology (fixed-speed instead of fixed-lift).
Here are the XFLR5 results for fixed-lift tests (using 0.4 kg) on the following designs:

Baseline: your original wing ($AR=5.0, \;S=0.05\;m^2, MAC=1.0$)
Tapered: your original tapered variant ($AR=6.67, \;S=0.038\;m^2, MAC=0.78$)
Tapered - equalized AR: the tapered variant with the root chord increased to match the aspect ratio of the Baseline ($AR=5.0, \;S=0.05\;m^2, MAC=1.08$)
Tapered - equalized area: the tapered variant with the span extended to match the area of the Baseline ($AR=8.8, \;S=0.05\;m^2, MAC=0.78$)

As expected, efficiency gains are proportional to AR increases. You can have a look at the updated project file here (make sure to click Show all Polars).

This is it i believe, I used fixed speed instead of fixed lift. I still have a couple semesters until my aero classes. I'm guessing we use fixed lift comparing the two airframes because they have their own optimized flight window right? i.e., the higher AR aircraft would want to fly a little faster to perform better, etc.

It may be inaccuracies in your modeling method; it's hard to say without more info.
First, this isn't quite an apples-to-apples comparison. The aspect ratio of the first wing is higher than the second wing, so you're varying more parameters than just the tip chord.
The way to think about this is that the CD of a wing is going to be strongly dependent on the aspect ratio $AR$, the taper ratio $\lambda_T$, the chord and twist distribution along the wing, as well as the airfoil distribution. Is the airfoil t/c constant across both wings? That will also have an effect.
It would be clearer to look at a polar of CL plotted against CD; the best point of comparison is to look at the drag generated at the same CL, not the same angle of attack (because the zero-lift angle of attack will change with wing design as well). The second wing is generating less drag, but also less lift at a given angle of attack; the comparison isn't really fair. If you're using a VLM what you're measuring is the CDi of a wing geometry, which is going to be most strongly dependent on CL.

The aspect ratio is higher, so why is there more drag? Shouldnt a higher aspect ratio lead to less induced drag if the wingspan is the same? You have the same parasitic drag if the wingspan is the same, etc. I'm definitely starating to get it now though with the CL vs CD graph, but the thicker wing still definitely excels there. The only graph where the thinner wing does better is CL/CD vs Alpha, which kind of clicks? The airfoil is the same t/c in both of the aircrafts (i'm assuming this is thickness % at chord of airfoil)

One more graph is the Moment vs Alpha graph, where the thicker wing is generating MUCH less CM at any given alpha, and the thinner wing is generating negative/stable CM. Could this also lead to the higher cd?

The right way to compare the two wings is looking at the CD values of the wings at the same CL, not the same alpha. So if that bottom plot is indeed CL vs. CD, you can see that the higher aspect ratio wing has lower drag at the same CL; this is what you would expect. Typically in flight alpha will be adjusted to give the target CL (and hence airspeed); the specific alpha value is not that significant.   Pitching moment doesn't have an effect on drag.

Thats what was confusing me so much. The XFLR5 is the complete opposite of that expectation, the lower AR wing has higher CL at the same CD. I messed up the data on the google sheets CL^2/CD graph that i made (wrong CD values for LowAR wing). It makes sense that this doesnt make sense right?

Does your method show the breakdown between viscous drag ($C_{D_{p}}$) and induced drag ($C_{D_{i}}$)? From the screenshot it looks like you're running a viscous analysis (coupled to a vortex lattice?). If you're only looking at $C_{D_{i}}$, then higher AR should give lower drag.  If you're including $C_{D_{p}}$ there may be more going on.

I just checked this as well, viscous drag was higher but induced drag (as expected) was lower. It seems that my method for the analysis (Fixed speed instead of fixed lift) was the issue here.


text:
Can drones/UAVs fly over international waters without being subject to the FAA's regulations?
Can UAVs or drones (consumer/commercial scale) operate over international waters (12 miles out from the shore) without being regulated by the FAA?
Just wondering if companies can do things like BVLOS* testing out in international waters if they wanted to, instead of getting exemptions or leasing private airspace.
And, this is meant to be living and residing in the US, and operating outside of US' waters.
* Beyond Visual Line Of Sight

At sea you'd have to fly your drone over the horizon to get it out of sight - at least 3 miles - and then there are no obstructions around which you can fly. Not a great test, I think.

I’m voting to close this question because this is a legal question and belongs on Law.SE.

@Ron I disagree. There are many questions here about the legalities of aviation-related  activities

Also, you'd have to take-off and land outside the ADIZ zone surrounding the US, otherwise you'd need a DVFR as well (although filling out APIS would be interesting).

This might be a question to post to the new [drones.se], still in private beta.

I'll go ahead and ask there but would like to keep it up here as well since there's more traffic here. FAA part 107 says visual line of sight is being able to see the aircraft without any aided sight, so even outside of a mile range might be considered BVLOS. Say you took off outside ADIZ zone though, you're free of the major bulk of regulation like 107? ADIZ zone extends pretty far but it's thin on the south edge, like near florida or southwest california

I assume you are based in the USA and are operating out of the USA? Please add that information to the question.

I dont suppose FAA's regulations would apply in brazil now. I'll add it though :)

Short answer: the FAA's UAV regulations probably stop applying at 3 nautical miles, but this is more about maritime law than aviation regulations and you might get a better answer on law.SE.

The FAA's UAV regulations do state where they apply. 14 CFR 107.1(a) (emphasis mine):

Except as provided in paragraph (b) of this section, this part applies
  to the registration, airman certification, and operation of civil
  small unmanned aircraft systems within the United States

14 CFR 1.1:

United States, in a geographical sense, means (1) the States, the District of Columbia, Puerto Rico, and the possessions, including the
  territorial waters, and (2) the airspace of those areas

Defining different types of marine jurisdiction is apparently so complicated that it requires a diagram (33 CFR 2.1), and the definition of "territorial sea" is in 33 CFR 2.22. Unfortunately, there are two possible definitions: 12 nautical miles or 3 nautical miles from shore. My reading of it is that for treaties, criminal law and other 'big stuff', the waters extend 12 miles. For more general purposes, it's 3 miles per 2.22(a)(2):

Unless otherwise specified in paragraph (a)(1) of this section,
  territorial sea means the waters, 3 nautical miles wide, adjacent to
  the coast of the United States and seaward of the territorial sea
  baseline.

Having said all that, I don't know for sure what the definition of "US territorial waters" really is. The 3NM distance seems very likely (to me) for part 107, and you might get a better answer on law.SE.


text:
How to join Air Force Reserve with minor injuries that would interfere with Basic Training?
Is it possible to get waivers on fitness part of the Basic Training for joining to Air Force Reserve? If not, is there a way to join Air Force Reserves without Basic Training?
Are there alternative paths that would skip Basic Training, for example: Civil Air Patrol (CAP) allows senior members and does not have fitness requirements as I know it, is there any way to transition to Air Force from there?
This does not need to just focus on flying. Although the pilot in question is already a commercial pilot, and has first class medical, just has some injuries to prevent extreme sports activities. (Like running 1.5 miles in 12 mins) So being able to fly would be very desirable (while not necessary).

Which fitness requirements exactly? Pilots are highly coveted positions, most are taken through military academies or ROTC programs. There is no transition from CAP to pilot program, and almost no waivers are allowed (I speak from great experience).

CAP and the military are two different things. CAP, Eagle Scouts, and ROTC might give you a leg up on being promoted from a private upon enlistment. But, it’s not going to give you much of an advantage past that.  If you want to be a pilot in the military, you have 2 routes. Either get a degree and become an officer. Or, join as enlisted in any capacity. Then, earn the opportunity (after a few years) to apply to warrant officer training for a rotary wing component. Either way, get your PT squared away now.

If you have a physical or mental issue that might keep you out of the military, your other option is to go to a flight school or collegiate flight program. Earning your flight certificate will give you a little credit toward OCS or WOCS. If not, at least you can be a civilian pilot. If you can not at least get a third class medical certificate on the civilian side, kiss your military flight dreams goodbye.

I updated the question. Question is not about how to become a pilot but more about how to join Air Force Reserves and serve (in any position) without fitness requirements. (Although being pilot would be great too)

Willingness to grant waivers depends entirely on the needs of the service, and this varies depending on factors like budget, retention, etc.  Only a recruiter will have the latest information.  Talk to the USAF, not random internet keyboard jockeys like us.

The Air Force used to grant waivers for the PT test if you could pass an alternate event. For instance, a timed bike ride instead of running to test your cardiovascular health. These waivers were usually granted to members already in the AF branch. Members going through basic training were still required to pass the regular PT test. The AF branch consists of Active, Reserve, and Air National Guard. All of which require their members to go through the same basic training together. CAP is an auxiliary of the AF. It is not technically in the AF branch. There is no easy route to AF through them.

@Dean F. - Yes, excellent point.  Waivers are generally for those who have paid their dues and the service has invested time and training in, and are later injured or have some condition that presents a challenge.  Why would they waiver an unproven person if they have a pool of qualified applicants to choose from?  Sucks for your friend, but life isn't fair.

The aviation medical requirements are strict. Flying a combat plane is physically demanding on it's own but imagine a scenario of ejecting over the ocean or in a wilderness area and having to survive. Couple this with the fact they choose the aircraft you fly, not you. That means you cant hide in a cargo plane if that isn't where they want you. Even if you don't want to be a pilot, the physical side of basics is designed to test you mental toughness and teamwork under pressure. You don't just get waivers for that because they need to assess if you have what it takes.

If you happen to be an expert in some particular field that the military thinks it has a critical need for, you are more likely to get a waiver.  I've seen this happen with certain doctors who were specialists that the military had a critical shortage of for whatever reason.  These were not people just out of med school though.  They were senior specialists who needed waivers for age at the very least.  I'm not sure such a situation would ever occur for a pilot though since there are plenty of pilots out there who figure that they could "fly the box the plane came in", regardless of the type of plane.

Good point, Grumpy. I have seen the military give entry waivers to high level, uniquely skilled experts in a much needed field. In recent times, the military has been going the civilian consultant or contractor route to fill these positions. This keeps the individual out of the military leadership career tract. I’ve seen this with doctors, cyber security, & drone pilots. More often, I see waivers for higher ranking enlisted & officers (E-8, O-5 & above) regardless of their specialty. I would not count on any waivers below E-5 unless you have a mid to high six figure civilian salary specialty.